Question: Simplify the following expression: $y = \dfrac{8x^2+37x+20}{8x + 5}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(8)}{(20)} &=& 160 \\ {a} + {b} &=& &=& {37} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $160$ and add them together. The factors that add up to ${37}$ will be your ${a}$ and ${b}$ When ${a}$ is ${5}$ and ${b}$ is ${32}$ $ \begin{eqnarray} {ab} &=& ({5})({32}) &=& 160 \\ {a} + {b} &=& {5} + {32} &=& 37 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({8}x^2 +{5}x) + ({32}x +{20}) $ Factor out the common factors: $ x(8x + 5) + 4(8x + 5)$ Now factor out $(8x + 5)$ $ (8x + 5)(x + 4)$ The original expression can therefore be written: $ \dfrac{(8x + 5)(x + 4)}{8x + 5}$ We are dividing by $8x + 5$ , so $8x + 5 \neq 0$ Therefore, $x \neq -\frac{5}{8}$ This leaves us with $x + 4; x \neq -\frac{5}{8}$.